Question: What is the extraneous solution to these equations? $\dfrac{x^2 + 14}{x - 7} = \dfrac{12x - 21}{x - 7}$
Multiply both sides by $x - 7$ $ \dfrac{x^2 + 14}{x - 7} (x - 7) = \dfrac{12x - 21}{x - 7} (x - 7)$ $ x^2 + 14 = 12x - 21$ Subtract $12x - 21$ from both sides: $ x^2 + 14 - (12x - 21) = 12x - 21 - (12x - 21)$ $ x^2 + 14 - 12x + 21 = 0$ $ x^2 + 35 - 12x = 0$ Factor the expression: $ (x - 5)(x - 7) = 0$ Therefore $x = 5$ or $x = 7$ At $x = 7$ , the denominator of the original expression is 0. Since the expression is undefined at $x = 7$, it is an extraneous solution.